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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
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tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /max_sum_circular_subarr.c
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+// 918. Maximum Sum Circular Subarray
+#include <stdio.h>
+#define MAX(a, b) (((a) > (b)) ? (a) : (b))
+
+/*
+ * given: circular int array 'nums' of length 'n', return the max possible
+ * sum of non-empty subarray of 'nums'
+ *
+ * a circular array means the end of the array connects to the beginning.
+ * formally, the previous element of 'nums[i]' is 'nums[(i -/+ 1) % n]'
+ *
+ * a subarray may only include each element of the fixed buffer 'nums' at
+ * most once. formally, for a subarray 'nums[i], nums[i + 1], ..., nums[j]'
+ * there does not exist i <= k1, k2 <= j with k1 % n == k2 % n
+ */
+
+int helper(int *a, int n) {
+ // kadane's dp algorithm
+ // [https://medium.com/@rsinghal757/kadanes-algorithm-dynamic-programming-how-and-why-does-it-work-3fd8849ed73d]
+ int currMax, sum;
+ sum = currMax = a[0];
+ for (int i = 1; i < n; i++) {
+ currMax = MAX(a[i], currMax + a[i]);
+ sum = MAX(currMax, sum);
+ }
+ return sum;
+}
+
+int maxSubarraySumCircular(int *nums, int numsSize) {
+ int k1, k2, total = 0;
+ k1 = helper(nums, numsSize);
+ for (int i = 0; i < numsSize; i++) {
+ total += nums[i];
+ nums[i] *= -1;
+ }
+ k2 = helper(nums, numsSize);
+ int ans = total + k2;
+ return (ans > k1 && ans != 0) ? ans : k1;
+}
+
+int main() {
+ int nums1[] = {1, -2, 3, -2}, nums2[] = {5, -3, 5}, nums3[] = {-3, -2, -3};
+ printf("%d\n", maxSubarraySumCircular(nums1, 4)); // expect: 3
+ printf("%d\n", maxSubarraySumCircular(nums2, 3)); // expect: 3
+ printf("%d\n", maxSubarraySumCircular(nums3, 3)); // expect: 3
+}